3.93.88 \(\int \frac {4 x^2+e^{\frac {e^{\frac {1}{4} (20-24 x-5 x \log (x))}}{x}+\frac {1}{4} (20-24 x-5 x \log (x))} (-4-29 x-5 x \log (x))}{4 x^2} \, dx\)

Optimal. Leaf size=24 \[ e^{\frac {e^{5-x-\frac {5}{4} x (4+\log (x))}}{x}}+x \]

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Rubi [F]  time = 0.75, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {4 x^2+\exp \left (\frac {e^{\frac {1}{4} (20-24 x-5 x \log (x))}}{x}+\frac {1}{4} (20-24 x-5 x \log (x))\right ) (-4-29 x-5 x \log (x))}{4 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4*x^2 + E^(E^((20 - 24*x - 5*x*Log[x])/4)/x + (20 - 24*x - 5*x*Log[x])/4)*(-4 - 29*x - 5*x*Log[x]))/(4*x^
2),x]

[Out]

x - Defer[Int][E^(5 - 6*x + E^(5 - 6*x)*x^(-1 - (5*x)/4))*x^(-2 - (5*x)/4), x] - (29*Defer[Int][E^(5 - 6*x + E
^(5 - 6*x)*x^(-1 - (5*x)/4))*x^(-1 - (5*x)/4), x])/4 - (5*Log[x]*Defer[Int][E^(5 - 6*x + E^(5 - 6*x)*x^(-1 - (
5*x)/4))*x^(-1 - (5*x)/4), x])/4 + (5*Defer[Int][Defer[Int][E^(5 - 6*x + E^(5 - 6*x)*x^(-1 - (5*x)/4))*x^(-1 -
(5*x)/4), x]/x, x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {4 x^2+\exp \left (\frac {e^{\frac {1}{4} (20-24 x-5 x \log (x))}}{x}+\frac {1}{4} (20-24 x-5 x \log (x))\right ) (-4-29 x-5 x \log (x))}{x^2} \, dx\\ &=\frac {1}{4} \int \left (4-e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-2-\frac {5 x}{4}} (4+29 x+5 x \log (x))\right ) \, dx\\ &=x-\frac {1}{4} \int e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-2-\frac {5 x}{4}} (4+29 x+5 x \log (x)) \, dx\\ &=x-\frac {1}{4} \int \left (4 e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-2-\frac {5 x}{4}}+29 e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-1-\frac {5 x}{4}}+5 e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-1-\frac {5 x}{4}} \log (x)\right ) \, dx\\ &=x-\frac {5}{4} \int e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-1-\frac {5 x}{4}} \log (x) \, dx-\frac {29}{4} \int e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-1-\frac {5 x}{4}} \, dx-\int e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-2-\frac {5 x}{4}} \, dx\\ &=x+\frac {5}{4} \int \frac {\int e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-1-\frac {5 x}{4}} \, dx}{x} \, dx-\frac {29}{4} \int e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-1-\frac {5 x}{4}} \, dx-\frac {1}{4} (5 \log (x)) \int e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-1-\frac {5 x}{4}} \, dx-\int e^{5-6 x+e^{5-6 x} x^{-1-\frac {5 x}{4}}} x^{-2-\frac {5 x}{4}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 21, normalized size = 0.88 \begin {gather*} e^{e^{5-6 x} x^{-1-\frac {5 x}{4}}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + E^(E^((20 - 24*x - 5*x*Log[x])/4)/x + (20 - 24*x - 5*x*Log[x])/4)*(-4 - 29*x - 5*x*Log[x]))
/(4*x^2),x]

[Out]

E^(E^(5 - 6*x)*x^(-1 - (5*x)/4)) + x

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fricas [B]  time = 0.58, size = 61, normalized size = 2.54 \begin {gather*} {\left (x e^{\left (-\frac {5}{4} \, x \log \relax (x) - 6 \, x + 5\right )} + e^{\left (-\frac {5 \, x^{2} \log \relax (x) + 24 \, x^{2} - 20 \, x - 4 \, e^{\left (-\frac {5}{4} \, x \log \relax (x) - 6 \, x + 5\right )}}{4 \, x}\right )}\right )} e^{\left (\frac {5}{4} \, x \log \relax (x) + 6 \, x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x*log(x)-29*x-4)*exp(-5/4*x*log(x)-6*x+5)*exp(exp(-5/4*x*log(x)-6*x+5)/x)+4*x^2)/x^2,x, alg
orithm="fricas")

[Out]

(x*e^(-5/4*x*log(x) - 6*x + 5) + e^(-1/4*(5*x^2*log(x) + 24*x^2 - 20*x - 4*e^(-5/4*x*log(x) - 6*x + 5))/x))*e^
(5/4*x*log(x) + 6*x - 5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, x^{2} - {\left (5 \, x \log \relax (x) + 29 \, x + 4\right )} e^{\left (-\frac {5}{4} \, x \log \relax (x) - 6 \, x + \frac {e^{\left (-\frac {5}{4} \, x \log \relax (x) - 6 \, x + 5\right )}}{x} + 5\right )}}{4 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x*log(x)-29*x-4)*exp(-5/4*x*log(x)-6*x+5)*exp(exp(-5/4*x*log(x)-6*x+5)/x)+4*x^2)/x^2,x, alg
orithm="giac")

[Out]

integrate(1/4*(4*x^2 - (5*x*log(x) + 29*x + 4)*e^(-5/4*x*log(x) - 6*x + e^(-5/4*x*log(x) - 6*x + 5)/x + 5))/x^
2, x)

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maple [A]  time = 0.08, size = 19, normalized size = 0.79

method	result	size
risch	\(x +{\mathrm e}^{\frac {x^{-\frac {5 x}{4}} {\mathrm e}^{5-6 x}}{x}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-5*x*ln(x)-29*x-4)*exp(-5/4*x*ln(x)-6*x+5)*exp(exp(-5/4*x*ln(x)-6*x+5)/x)+4*x^2)/x^2,x,method=_RETUR
NVERBOSE)

[Out]

x+exp(x^(-5/4*x)*exp(5-6*x)/x)

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maxima [A]  time = 0.60, size = 18, normalized size = 0.75 \begin {gather*} x + e^{\left (\frac {e^{\left (-\frac {5}{4} \, x \log \relax (x) - 6 \, x + 5\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x*log(x)-29*x-4)*exp(-5/4*x*log(x)-6*x+5)*exp(exp(-5/4*x*log(x)-6*x+5)/x)+4*x^2)/x^2,x, alg
orithm="maxima")

[Out]

x + e^(e^(-5/4*x*log(x) - 6*x + 5)/x)

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mupad [B]  time = 7.61, size = 20, normalized size = 0.83 \begin {gather*} x+{\mathrm {e}}^{\frac {{\mathrm {e}}^{-6\,x}\,{\mathrm {e}}^5}{x^{\frac {5\,x}{4}}\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - (exp(5 - (5*x*log(x))/4 - 6*x)*exp(exp(5 - (5*x*log(x))/4 - 6*x)/x)*(29*x + 5*x*log(x) + 4))/4)/x^2
,x)

[Out]

x + exp((exp(-6*x)*exp(5))/(x^((5*x)/4)*x))

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sympy [A]  time = 0.68, size = 19, normalized size = 0.79 \begin {gather*} x + e^{\frac {e^{- \frac {5 x \log {\relax (x )}}{4} - 6 x + 5}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x*ln(x)-29*x-4)*exp(-5/4*x*ln(x)-6*x+5)*exp(exp(-5/4*x*ln(x)-6*x+5)/x)+4*x**2)/x**2,x)

[Out]

x + exp(exp(-5*x*log(x)/4 - 6*x + 5)/x)

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